leetcode——107——Binary Tree Level Order Traversal II

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int>> res;
		 if (root == NULL) return res;
		 queue<TreeNode*>q;
		 q.push(root);
		 while (!q.empty()){
			 vector<int> ans;
			 int size = q.size();
			 for (int i = 0; i < size; i++){
				 TreeNode *temp = q.front();
				 q.pop();
				 ans.push_back(temp->val);
				 if (temp->left)
					 q.push(temp->left);
				 if (temp->right)
					 q.push(temp->right);
			 }
			 res.push_back(ans);
		 }
         reverse(res.begin(), res.end());
		 return res; 
    }
};